vector spaces, dimension
F-111
over F. (N = the naturals).
Is it true that Dim V >= Card F ?
Gerry Myerson
<1285344499.38026.12591...@gallium.mathforum.org>,
F-111 <tv_...@live.com> wrote:
> Let F be some abelian field,
What other kind of field is there?
> and let V := F^N, considered as a vector space
> over F. (N = the naturals).
> Is it true that Dim V >= Card F ?
It seems to me that the dimension of V is independent
of F, so if there are fields of arbitrarily large cardinality,
the inequality should fail.
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
Virgil
Are there any fields which are not abelian?
Do you mean a division ring (or skew field)?
Tim Little
> It seems to me that the dimension of V is independent of F, so if
> there are fields of arbitrarily large cardinality, the inequality
> should fail.
ZFC permits ultrapower fields of unbounded cardinality, but I'm not
certain that dim V is independent of card F. For very large F, I
don't see how every infinite sequence could be formed with finite
linear combinations from only continuum many basis elements.
I suspect that for |F| >= |2^N|, dim V = |F|.
- Tim
A N Niel
Virgil <Vir...@home.esc> wrote:
I think that is from Bourbaki ... translated to English, "field" and
"commutative field".
Ken Pledger
You say V = F^N, considered as a vector space over F.
Therefore Dim V = the cardinality of N,
so any uncountable field F (e.g. the real or complex field) will have
greater cardinality.
Ken Pledger.
F-111
> <1285344499.38026.1259195146096.JavaMail.root@gallium.
> mathforum.org>,
> F-111 <tv_...@live.com> wrote:
>
> > Let F be some abelian field, and let V := F^N,
> considered as a vector space
> > over F. (N = the naturals).
> > Is it true that Dim V >= Card F ?
>
>
> You say V = F^N, considered as a vector space
> space over F.
> Therefore Dim V = the cardinality of N,
No, no. Generally, for any two sets A, B one denote
A^B := {f | f:B -> A}.
So that, in our case, V is the set of ALL F-valued sequences.
Arturo Magidin
> In article
> <1285344499.38026.1259195146096.JavaMail.r...@gallium.mathforum.org>,
>
> F-111 <tv_...@live.com> wrote:
> > Let F be some abelian field, and let V := F^N, considered as a vector space
> > over F. (N = the naturals).
> > Is it true that Dim V >= Card F ?
>
> You say V = F^N, considered as a vector space over F.
> Therefore Dim V = the cardinality of N,
This does not hold; for example, if F=Q, then Q^N is uncountable as a
set, and therefore must have an uncountable basis; a countable set
spans only a countable set.
--
Arturo Magidin
Gerry Myerson
Tim Little <t...@little-possums.net> wrote:
Looks like I didn't think this through. Can you find a proof
or accessible reference?
Tim Little
>> I suspect that for |F| >= |2^N|, dim V = |F|.
>
> Looks like I didn't think this through. Can you find a proof
> or accessible reference?
No, since it's just a suspicion. I can't prove it at the moment, and
perhaps it is in fact true that dim V is independent of |F|. I just
have an intuition that if dim V = |R| despite |F| > |R|, then the
basis can only employ a "tiny" subset of elements of F, and would have
to stretch only finitely many coefficients from the full field to
somehow generate all the infinite sequences.
Unfortunately a proof that this cannot be done is not forthcoming.
- Tim
David Bernier
For a general field F, can we at least show
that dim(F^N) > aleph_0 ?
David Bernier
David C. Ullrich
No. If F is a field and S is an infinite set then card(S) is the
dimension of a certain subspace V of F^S, namely the set
of elements that equal 0 in all but finitely many coordinates.
V is typically a very small subspace of F^S.
> Ken Pledger.
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
Tim Little
> For a general field F, can we at least show that dim(F^N) > aleph_0?
It's easy for countable F, since a countable set of vectors with
countable coefficients can only span a countable subspace. In
particular, {0,1}^N has uncountable basis.
Then show that for uncountable F, any independent set in {0,1}^N can
be mapped to an independent set in F^N, so its basis is at least as
large.
- Tim
Arturo Magidin
I think you can use tensor products; take K to be the prime field of
F, and look at K^N. This has dimension c = 2^{aleph_0}. So K^N is the
direct sum of c copies of K. Since tensor squares commute with direct
sums, the tensor with F, considered as a vector space over F, will
have a basis of the form b_i(x)1, where {b_i} is the basis of K^N.
Now comes the sticky point: it is not clear to me whether K^N(x)F =
F^N. It should certainly map into F^N giving the inequality desired;
but in general, tensors commute with direct sums, not arbitrary direct
products, so I don't know if we get equality or just inclusion (or if
I'm totally off base here).
--
Arturo Magidin
F-111
Since |F| is just the dimension of W, i.e., the F-vector space of all
functions w: F --> F having finite support, it would be enough to
define a "total" sequence {f_n} of F-linear mappings f_n : W --> F.
By "total" I mean "f_n(w) = 0 for all n in N ==> w = 0". That's in order to
obtain a F-linear injective map L: W --> V, Lw := {f_n(w)}.
A remarkable "functional" f:W --> F is f(w) := Sum { w(f) ; f in F }.
That's all for now.
F-111
> In article
> <1285344499.38026.1259195146096.JavaMail.root@gallium.
> mathforum.org>,
> F-111 <tv_...@live.com> wrote:
>
> > Let F be some abelian field,
>
> What other kind of field is there?
"Theory of commutative fields" by M.Nagata, Transl. AMS, 1993.
>
> > and let V := F^N, considered as a vector space
> > over F. (N = the naturals).
> > Is it true that Dim V >= Card F ?
>
> It seems to me that the dimension of V is independent
>
> of F, so if there are fields of arbitrarily large
> cardinality,
> the inequality should fail.
Actually, F^N and F[[X]] are isomorphic as F-vector spaces.